Problem: Divide the following complex numbers. $\dfrac{6-7i}{4+i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${4-i}$. $ \dfrac{6-7i}{4+i} = \dfrac{6-7i}{4+i} \cdot \dfrac{{4-i}}{{4-i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(6-7i) \cdot (4-i)} {4^2 - (i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(6-7i) \cdot (4-i)} {(4)^2 - (i)^2} $ $ = \dfrac{(6-7i) \cdot (4-i)} {16 + 1} $ $ = \dfrac{(6-7i) \cdot (4-i)} {17} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({6-7i}) \cdot ({4-i})} {17} $ $ = \dfrac{{6} \cdot {4} + {-7} \cdot {4 i} + {6} \cdot {-1 i} + {-7} \cdot {-1 i^2}} {17} $ $ = \dfrac{24 - 28i - 6i + 7 i^2} {17} $ Finally, simplify the fraction. $ \dfrac{24 - 28i - 6i - 7} {17} = \dfrac{17 - 34i} {17} = 1-2i $